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One needs to choose six real numbers $x_1,x_2,\cdots,x_6$ such that the product of any five of them is equal to other number. The number of such choices is

A) $3$

B) $33$

C) $63$

D) $93$

I believe this is not so hard problem but I got no clue to proceed. The work I did till now.

Say the numbers be $a,b,c,d,e, abcde$. Then, $b\cdot c\cdot d\cdot e\cdot abcde=a$ hence $bcde= +-1$.

Basically I couldn't even find a single occasion where such things occcur except all of these numbers being either $1$ or $-1$. Are these all cases?

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    $\begingroup$ They could all be $0$ (unless the image, which I can't access right now, has a rule forbidding that). $\endgroup$ – J.G. Mar 4 at 7:38
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    $\begingroup$ I have typed out your image - pleas do this in the future so users such as @J.G. can view your whole question $\endgroup$ – lioness99a Mar 4 at 11:13
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    $\begingroup$ For extra credit: how many complex numbers? {i,i,i,i,i,i,i} is one solution. Heck, how many quaternions?. $\endgroup$ – Carl Witthoft Mar 4 at 13:42
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    $\begingroup$ Seems related to the knapsack problem. $\endgroup$ – EJoshuaS Mar 5 at 16:48
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    $\begingroup$ In a similar way to @J.G. they could all be 1. $\endgroup$ – Andrew McOlash Mar 5 at 19:07
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Well, if there are an even number of $1$'s and $-1$'s, then the property holds, so that's $$\binom{6}{0}+ \binom{6}{2}+ \binom{6}{4}+ \binom{6}{6}=32$$ And then there’s the case that they’re all zero. Thus, there are $33$ total cases.

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  • $\begingroup$ I was in right direction then. These two last options just confused me. I was thinking there could be much more elements. $\endgroup$ – ChakSayantan Mar 4 at 7:47
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    $\begingroup$ You have made no argument as to why this accounts for all cases... $\endgroup$ – Morgan Rogers Mar 4 at 13:58
  • $\begingroup$ As Morgan points out, this is fine and all, but this is not a proof. This rules out option A, leaves the others open. $\endgroup$ – Clement C. Mar 5 at 17:07
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    $\begingroup$ @IMil, this is true, they were just highlighting that this answer could be more rigorous, which is simply true. the person who answered after me did things much much better in my opinion. $\endgroup$ – Zachary Hunter Mar 6 at 1:08
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    $\begingroup$ @ZacharyHunter sure. I also missed the next obvious step that magnitude has to be either 1 or 0 because otherwise the product will be either more or less than any of the numbers. $\endgroup$ – IMil Mar 6 at 1:13
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Here's an argument which extends to $n$ real numbers quite nicely! We can start by noting that $$x_1 (x_2x_3x_4x_5x_6) = x_1^2$$ And by commutativity, we get $x_i^2 = x_j^2$ which implies that all the magnitudes are equal.

Now if $L$ is the magnitude, we also must have $L = L^5$, and from this we conclude that $L=0$ or $1$.

Now we just have to go through the possibilities. If $L = 0$, then we have all $0$'s

If $L = 1$, then we have all $-1$'s and $1$'s. This configuration works iff the number of $-1$'s is even, as this would imply that $$\frac{\prod_{i=1}^6 x_i}{1} = 1 \text{ and } \frac{\prod_{i=1}^6 x_i}{-1} = -1$$ Now we can count configurations. There will be $$\sum_{i=0}^3 \binom{6}{2i} = 2^5$$ possibilities. And finally, we have $1+32 = 33$.

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    $\begingroup$ the 2n+1 real number case is pretty much the same, you just add all the cases where there are an even number of “-1”s. so I guess generally it’s $\sum_{i=0}^n \binom{n}{2i}$ $\endgroup$ – Zachary Hunter Mar 4 at 17:08
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    $\begingroup$ Good point! I'm not sure why I restricted myself to $2n$. Anyways, I'll go and replace $2n$ with $n$ now. $\endgroup$ – Isaac Browne Mar 4 at 21:03
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    $\begingroup$ This might be clearer if you point out that $x_i^2 = x_j^2$ for all $i$ and $j$ instead of jumping right to "by commutativity". I'm not certain why you called $32$ $2^5$ in your 2nd last line; are you claiming there is an identity easier than $6+10+10+6 = 32$ there? $\endgroup$ – Yakk Mar 5 at 2:16
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    $\begingroup$ @Yakk As for the former complaint, I'll try to edit that to make it more clear. As for the latter complaint, I wrote $2^5$ because in general we have $$\sum_{0 \leq 2i \leq n} \binom{n}{2i} = 2^{n-1}$$ This identity is why the argument generalizes so nicely! $\endgroup$ – Isaac Browne Mar 5 at 2:33
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    $\begingroup$ It’s easy to see that the number of subsets of an $n$-set with an even number of elements is always $2^{n-1}$: just note that the operation of flipping the inclusion of one fixed element is an involution between the odd-sized sets and the even-sized sets. $\endgroup$ – Erick Wong Mar 5 at 2:43
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Just to make the solution more explicit, the product of all six values is the square of any one of them, so they're all $\pm k$ for some $k$, and $k^6=k^2\implies k\in\{0,\,1\}$. We get one $k=0$ solution and, because we require a non-negative product, one $k=1$ solution for each choice of six $\pm 1$s with an even number of $-1$s. Famously, the numbers of even- and odd-sized subsets of a finite non-empty set are equal, and so the total number of solutions is $1+\frac12\cdot2^6$, i.e. B).

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Even though the question stated real numbers, we can extend this to complex fairly easily.

Note we require

$$ \exp( \left( E - I) \boldsymbol{z} \right ) = \exp(\boldsymbol{z}) $$

where $\exp(\boldsymbol{z}) = \boldsymbol{x}$ (the vector of numbers to multiply), $E$ is a matrix of ones and $I$ is the identity matrix.

Now,

$$ \exp( \left( E - 2I) \boldsymbol{z} \right ) = \exp(2\pi i \boldsymbol{k}) $$

for some vector of integers $\boldsymbol{k}$. Equating the powers and using $(E - 2I)^{-1} = \frac{1}{8}(E - 4I)$, gives

$$ \boldsymbol{z} = \frac{K \pi i}{4} - \pi i \boldsymbol{k}$$

where $K = \sum_{i=1}^6 (\boldsymbol{k})_i$ and hence

$$ \boldsymbol{x} = \exp \left (\frac{K \pi i}{4} \right) \cdot \exp(\pi i \boldsymbol{k})$$

The second term is clearly a vector of 1s and -1s. The first term can take the following values:

  1. If the total 1s in the second term are even i.e. (even number of even $k_i$),

    • $1$
    • $i$
    • $-1$
    • $-i$
  2. If the total 1s in the second term are odd i.e. (odd number of even $k_i$),

    • $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$
    • $-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$
    • $-(-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)$
    • $-(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)$

Finally, consider the subsets of $A = \{-1, 1\}^6$, $A_E \subset A$ containing an even number of 1s and $A_O \subset A$ containing an odd number of 1s.

Clearly $A_E = -A_E$ (multiplying each element in the set by -1) and $A_O = - A_O$, hence the unique combinations of complex numbers is the union of disjoint sets

$$C = A_E \cup iA_E \cup (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i) A_O \cup (-\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i) A_O$$

As $|A_E| = |A_O| = 32$, and the above is a union of disjoint subsets, $|C| = 128$ and including the case where all values are $0$, we have the total complex combinations as $129$.

Note

To answer the question, we know that the only subset containing real values is $A_E$ so the total combinations in the real case is $33$.

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HINT:

It is obvious that you need to have the numbers absolute value equal to $1$ (besides one special case but we get to that later). The question is how to set the signs of these $1$s. The product will be negative if you have an odd number of negative signs in it and positive whenever you have an even number of negative signs. Since you are to determine signs for an even number of $1$s if you decide to give an even number of them a negative sign you have one of the following two situations

  • the $1$ that is NOT in the product has negative sign. In this case there are a $\text{even}-1=\text{odd}$ number of negative signs distributed along the other five $1$s which result in $-1$ so the condition is fullfilled

  • the $1$ that in NOT in the product has positive sign. In this case there are a $\text{even}-0=\text{even}$ number of negative signs distributed along the five other $1$s which result in $+1$ so the condition is once again fullfilled.

So your question translates to in how many ways can you distribute an even number of negative signs among the six $1$s. This is solved in the easiest way by considering the number of ways to distribute $0,2,4$ and $6$ negative signs.

Finally the one solution not having to deal with $1$s is the one dealing with all zeros.

Hope this helped

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